Question: The equation of a circle $C$ is $x^2+y^2+6x+12y+20 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2+6x) + (y^2+12y) = -20$ $(x^2+6x+9) + (y^2+12y+36) = -20 + 9 + 36$ $(x+3)^{2} + (y+6)^{2} = 25 = 5^2$ Thus, $(h, k) = (-3, -6)$ and $r = 5$.